$f(x)=\lim_{h\to0}\frac{1}{h}\int\limits_{x}^{x+h}\frac{dt}{t+\sqrt{1+t^2}}$,then $\lim_{x\to-\infty}x.f(x)$ is Let $f(x)=\lim_{h\to0}\frac{1}{h}\int\limits_{x}^{x+h}\frac{dt}{t+\sqrt{1+t^2}}$,then $\lim_{x\to-\infty}x.f(x)$ is
$(A)0\hspace{1cm}(B)\frac{1}{2}\hspace{1cm}(C)1\hspace{1cm}(D)$non existent
My Attempt:
$f(x)=\lim_{h\to0}\frac{1}{h}\int\limits_{x}^{x+h}\frac{dt}{t+\sqrt{1+t^2}}$
This is a $\frac{0}{0}$ form,so i applied L'Hospital rule
$f(x)=\lim_{h\to0}\frac{1}{x+h+\sqrt{1+(x+h)^2}}-\frac{1}{x+\sqrt{1+x^2}}=0$
I am stuck now,dont know whether i am doing correct or not.Please help me in solving this question.
 A: From the Fundamental Theorem of Calculus, we have
$$\begin{align}
f(x)&=\lim_{h\to 0}\frac1h \int_x^{x+h}\frac{1}{t+\sqrt{1+t^2}}dt\\\\
&=\frac{1}{x+\sqrt{1+x^2}}
\end{align}$$
Therefore, 
$$\begin{align}
\lim_{x\to \infty}xf(x)&=\lim_{x\to \infty}\frac{x}{x+\sqrt{1+x^2}}\\\\
&=\frac12
\end{align}$$
So, then answer is $(B) \,\,\frac12$.
A: How did you apply L'Hospital? If there is an $h$ in the denominator, you should take the derivative with respect to $h$ in the numerator. I think you took the derivative with respect to $x$ instead. :)
Another way to do this: If we let $F(x) = \int\limits_{0}^{x}{\frac{dt}{t+\sqrt{1+t^2}}}$, then
\begin{align*} f(x) = \lim\limits_{h\rightarrow 0}{\frac{1}{h}\int\limits_{x}^{x+h}{\frac{dt}{t+\sqrt{1+t^2}}}} &= \lim\limits_{h\rightarrow 0}{\frac{1}{h}\left(\int\limits_{0}^{x+h}{\frac{dt}{t+\sqrt{1+t^2}}} - \int\limits_{0}^{x}{\frac{dt}{t+\sqrt{1+t^2}}}\right)} \\
&= \lim\limits_{h\rightarrow 0}{\frac{1}{h}(F(x+h)-F(x))} \\
&= F'(x) \end{align*}
Can you calculate what $F'(x)$ is?
A: Define $F(x) = \int_0^x \frac{1}{t + \sqrt{1 + t^2}} dt, x \in \mathbb{R}$.
The condition reads
$$f(x) = \lim_{h \to 0} \frac{F(x + h) - F(x)}{h}.$$
while the right hand side is the derivative of $F$ at $x$. Therefore 
$$f(x) = F'(x) = \frac{1}{x + \sqrt{1 + x^2}}.$$
Consequently,
$$\lim_{x \to -\infty} xf(x) = \frac{1}{1 + \frac{\sqrt{1 + x^2}}{x}} = \frac{1}{2}.$$
